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For $a,b>0$ are two real numbers and $p\geq 1$. Is the following inequality true $$|a^p-b^p|\leq|a-b|^p\;\;?$$

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  • $\begingroup$ 5=|3^2-2^2| > |3-2|^2=1 $\endgroup$ – ALG Oct 26 '18 at 21:10
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No, let $a=10,b=20,p=2$.

Then $|a^p-b^p|=|100-400|=300>100=|10-20|^2=|a-b|^p$.

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Let $b\geq a$ then $\>b=a+\delta\>$ and $\>\delta>0$ $$ \\|a^p-b^p|\leq|a-b|^p \\(a+\delta)^p-a^p\leq\delta^p \\(a+\delta)^p\leq a^p+\delta^p $$ but $a>0$ and $\delta\geq0 => (a+\delta)^p=a^p+\delta^p+\delta*x\>(x\geq0)($ Binomial theorem$)=>$ $$ \\a^p+\delta^p\geq(a+\delta)^p=a^p+\delta^p+\delta*x\geq a^p+\delta^p $$ $=>\delta*x=0=>\delta=0\>$ or $\>x=0=>$ $$ \\|a^p-b^p|\leq|a-b|^p $$ if and only if $$ a=b $$ or $$ p=1 $$

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Is

$$\left|1^2-\frac1{2^2}\right|\le\left|1-\frac12\right|^2\;?$$

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